Timeline for Regularity of a nonlinear ODE [Traveling wave solutions of parabolic systems]

6 events

when toggle format	what		by	license	comment
Oct 21, 2013 at 15:29	comment	added	leo monsaingeon		hehehe ;-) but I insist that the proof is not exactly the same when $\sigma<0$ and $\sigma>0$. In the first case you use De L'Hopital's rule, while in the second case $\sigma>0$ you find an equivalent of the divergent integral $\int\limits^t_{-\infty}e^{\sigma s}f(w(s))ds\underset{t\to\infty}{\sim}f(w_+)\frac{e^{\sigma t}}{\sigma}$ (even though this is still some kind of De L'Hopital's rule...)
Oct 21, 2013 at 15:13	history	edited	Kosh	CC BY-SA 3.0	A Remark and some explanation in the case $\sigma >0$.
Oct 21, 2013 at 14:43	comment	added	Kosh		Ok it is exactly what you have written :)
Oct 21, 2013 at 14:41	comment	added	Kosh		If $\sigma>0$ the proof is the same. Just let $t_0\rightarrow -\infty$ and then $t\rightarrow \pm \infty$
Oct 21, 2013 at 13:26	comment	added	leo monsaingeon		My observation $t\to-t\Leftrightarrow \sigma\to-\sigma$ absolutely doesn't imply that you can assume $\sigma<0$. It only means that if you can prove your statement on one side (say $t\to+\infty$) then the other side follows. Let's say you want to look at $t\to+\infty$: you still have to deal with both cases $\sigma>0$ and $\sigma<0$ so in (1) you should replace $-\|\sigma\|$ by just $\sigma$. I agree that your proof above works if $\sigma<0$ by letting $t_0\to\infty$ in (1). But if $\sigma>0$ sending $t_0\to\infty$ makes things diverge. Hint: if $\sigma>0$ rather let $t_0\to-\infty$!!
Oct 21, 2013 at 10:13	history	answered	Kosh	CC BY-SA 3.0